A survey of 60 randomly selected students finds that they spend a mean of $\$ 439$ per semester. Assume that the population standard deviation is $\$ 56$ per month.
Confidence Interval: What is the $90 \%$ confidence interval to estimate the population mean? Round your answers to two decimal places.
$< \mu< $
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Step 1 ：We are given that the sample mean (\(\bar{x}\)) is \$439, the Z-score (Z) for a 90% confidence level is 1.645, the population standard deviation (\(\sigma\)) is \$56, and the sample size (n) is 60.

Step 2 ：We can use these values to calculate the confidence interval for the population mean using the formula: \[\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\]

Step 3 ：Substituting the given values into the formula, we get: \[439 \pm 1.645 \frac{56}{\sqrt{60}}\]

Step 4 ：Solving this expression, we find the margin of error to be approximately 11.89.

Step 5 ：Subtracting this margin of error from the sample mean, we find the lower bound of the confidence interval to be \$427.11.

Step 6 ：Adding the margin of error to the sample mean, we find the upper bound of the confidence interval to be \$450.89.

Step 7 ：Therefore, the 90% confidence interval to estimate the population mean is \(\boxed{(\$427.11, \$450.89)}\).